Problem 3.8. Let (X, 𝒜) be a measurable space. In this...

Problem 3.8. Let $(X, \mathcal{A})$ be a measurable space. In this problem, we will show that $f, g \in \mathcal{M}_{\mathbb{R}}(\mathcal{A}) \implies f \cdot g \in \mathcal{M}_{\mathbb{R}}(\mathcal{A})$. (i) Prove that $f \in \mathcal{M}_{\mathbb{R}}(\mathcal{A}) \implies f^2 \in \mathcal{M}_{\mathbb{R}}(\mathcal{A})$. (ii) Use (i) to show that $f, g \in \mathcal{M}_{\mathbb{R}}(\mathcal{A}) \implies f \cdot g \in \mathcal{M}_{\mathbb{R}}(\mathcal{A})$.

Transcript

00:01 In this problem, our job is to show that if f and g are functions that are integrable on the closed bounded interval from a to b, then so is their product.

00:10 So let's make a summary first of the known results we have to help us here.

00:16 We know from the previous problem that if f is integrable, we'll just abbreviate that int, on this closed bounded interval from a to b, then f squared is also integrable on a, b.

00:32 That's the first result we'll need here.

00:34 Here.

00:39 Next, we will need the result that if f and g are functions that are integrable on that interval, then so are their sum and their difference.

00:57 So we'll just write that as f plus or minus g.

01:03 So we know that the sum and the difference of the two functions is also integrable.

01:10 And finally, we know that if f is integrable on that interval from a to b, and if c is any real number, any constant, then c times f is integrable on a, b.

01:30 So we're going to use these three results to prove that the product fg is also integrable.

01:40 We will use the hint that's provided as well.

01:44 So let's see how we can use these three results.

01:47 Make a little bit more room here.

01:52 So the first key observation, as the hint provides, is that f plus g squared as a function is the same as the function f squared plus 2fg plus g squared, and we notice that we have our goal right here.

02:14 So the idea is can we solve for fg and get some useful information? if we solve solve for fg, we will have 1 half times f plus g squared minus f squared minus g squared.

02:37 And now we're going to use the result above to show that this function is integrable.

02:41 So is this, so is this, and multiplying by a constant means that the result is integrable.

02:48 So we will apply each of those results in turn to this expression...

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